If y=a+b _e x then which of the following is true? Option 1: 1/(y-a) is proportional to x^b Option 2: y-a is proportional to x^b Option 3: e^y is proportional to x^b Option 4: _e y is proportional to x

Answer: e^y is proportional to x^b. Solution: Subtract a from both sides: y - a = b _e x (y - a)/b = _e x If _e x = k, then x = e^k So: x = e^((y - a)/b) Using exponent rules: x = e^((y - a)/b) = e^(y/b - a/b) = e^(y/b) * e^(-a/b) Since e^(-a/b) is a constant, let's call it k: x = k * e^(y/b) Raise both sides to the power b: x^b = k^b * e^y Since k^b is still a constant: x^b = constant x e^y Therefore: e^y is proportional to x^b

Modern Math

Logarithms

Medium

If $y=a+b \log _{e} x$ then which of the following is true?

$\frac{1}{y-a}$ is proportional to $x^{b}$

$y-a$ is proportional to $x^{b}$

$e^{y}$ is proportional to $x^{b}$

$\log _{e} y$ is proportional to $x$

✅ Correct Option: 3

Subtract $a$ from both sides:

$y - a = b \log_e x$

$\frac{y - a}{b} = \log_e x$

If $\log_e x = k$ , then $x = e^k$

So:

$x = e^{\frac{y - a}{b}}$

Using exponent rules:

$x = e^{\frac{y - a}{b}}$

$= e^{\frac{y}{b} - \frac{a}{b}}$

$= e^{\frac{y}{b}} \cdot e^{-\frac{a}{b}}$

Since $e^{-\frac{a}{b}}$ is a constant, let's call it $k$ :

$x = k \cdot e^{\frac{y}{b}}$

Raise both sides to the power $b$ :

$x^b = k^b \cdot e^y$

Since $k^b$ is still a constant:

$x^b = \text{constant} \times e^y$

Therefore: $e^y$ is proportional to $x^b$

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